Question

A truck heading due west increases its speed with a constant acceleration from 12 m/s to 18 m/s in 7.5 seconds. How far does the truck travel during this time? What was the acceleration of the truck during this period?

Answer 1

`acceleration=0.8 (m)/(s^(2) ) \\distance=112.5 m`

Step-by-step explanation:

We can find these answers following the equations of motion.

1. To find the acceleration, we use the equation:

`a=`Δ
`V/t`

Where Δ
`V` is the difference between the final speed and the initial speed. And
`t` is the time spent

We replace the terms:

`a=(18(m)/(s) -12(m)/(s) )/(7.5s)`

We solve the difference:

`a=(6(m)/(s))/(7.5s)`

We divide the terms, so we can have the answer:

`a=0.8 (m)/(s^(2))`

2. To find the distance traveled by the truck, we use the equation:

`x=V_0t+(1)/(2)at^(2)`

Where
`x` is the distance traveled,
`V_0` is the initial speed,
`a` is the acceleration and
`t` is the time.

We replace the terms:

`x=(12(m)/(s)*7.5s)+(1)/(2)[0.8(m)/(s^(2) )*(7.5)^(2) ]`

We multiply and solve the exponential:

`x=90m+(1)/(2)(0.8(m)/(s^(2) )*56.25s^(2) )`

Then, we multiply the terms left:

`x=90m+22.5m`

And add, so we can have the answer:

`x=112.5m`